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24 tháng 8 2019
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
24 tháng 8 2019
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
NH
10 tháng 4 2020
a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x-9=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
Vậy....
b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow-10+17x=-44\)
\(\Leftrightarrow17x=-34\)
\(\Leftrightarrow x=-2\)
Vậy....
c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy....
d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)
\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)
\(\Leftrightarrow14x-3=7\)
\(\Leftrightarrow14x=10\)
\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)
Vậy...
NM
Nguyễn Minh Quang
Giáo viên
4 tháng 8 2021
ta có
a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)
\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)
\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)
b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2
\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)
\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)
15 tháng 8 2022
a: \(\Leftrightarrow5x^2-45x-7x+63-\left(5x-2\right)\left(x-3\right)=2x^2-8x-2x^2-3x+2x+3\)
\(\Leftrightarrow5x^2-52x+63-\left(5x-2\right)\left(x-3\right)=-9x+3\)
\(\Leftrightarrow5x^2-52x+63-5x^2+15x+2x-6=-9x+3\)
=>-37x+57=-9x+3
=>28x=-54
hay x=-27/14
b: \(\Leftrightarrow-x^2+19x+3x-30+x^2-5x-24=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)
\(\Leftrightarrow17x-54=5x^3+15x^2-x-3-5x^3-15x^2\)
=>18x=51
hay x=17/6
PA
19 tháng 2 2017
a)
\(\frac{1}{x-2}+3=3-\frac{x}{x-2}\)
<=> \(\frac{1}{x-2}=-\frac{x}{x-2}\)
<=> x = - 1
Vậy S = {- 1}
b)
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
<=> \(\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)
<=> (x + 5)2 - (x - 5)2 = 20
<=> (x + 5 - x + 5)(x + 5 + x - 5) = 20
<=> 10 . 2x = 20
<=> x = 20 : 20
<=> x = 1
Vậy S = {1}
c)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{2\left(x-3\right)\left(x+1\right)}\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x}{2\left(x-3\right)\left(x+1\right)}\)
<=> x(x + 1) + x(x - 3) = 2x
<=> x2 + x + x2 - 3x - 2x = 0
<=> 2x2 - 4x = 0
<=> 2x(x - 2) = 0
<=> \(\left[\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy S = {0; 2}
19 tháng 2 2017
Bạn có sửa đề cũng phải báo chứ:
làm vậy có ai đó vào thấy đúng copy pas đến chỗ khác thành sai=> mất kiểm soát.
Tam sao thất bản mà.
Ngàn Sao thì ....
p/s: xem bài chứng tỏ bạn là đời f(0)
hiihi nói vui nhé xin đừng chém.
18 tháng 9 2020
(x+2)^2= 9
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5
18 tháng 9 2020
Bài làm :
\(a,\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Leftrightarrow\left(x+2\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x = 1 hoặc x = -5 .
\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy x = 2 .
\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3+x^2+x-x^2-x-1+\left(x^2+2x\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+4x=5+1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 .
Học tốt nhé
12 tháng 6 2018
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)